1.The rate of flow of electric charge...
Ans: Electric Current.
2.The SI unit of Current Density ($J$) is...
Ans: Ampere per square meter ($\text{A/m}^2$).
Ans: (B) Scalar quantity (It has magnitude and direction, but does not obey vector addition rules).
4.Calculate current if $20\text{ C}$ flows in $5\text{ s}$.
Ans: $I = \frac{q}{t} = \frac{20}{5} = 4\text{ A}$.
5.Define current density. Is it scalar or vector?
Ans: Current density at a point is the electric current flowing normally per unit area at that point. It is a vector quantity ($\vec{J}$).
6.Find $J$ for $I = 2\text{ A}$ and $A = 1\text{ mm}^2$.
Ans: $A = 1\text{ mm}^2 = 10^{-6}\text{ m}^2$. $J = \frac{I}{A} = \frac{2}{10^{-6}} = 2 \times 10^6\text{ A/m}^2$.
7.According to Ohm's law... proportional to the...
Ans: Potential difference (Voltage).
8.SI unit of electrical resistivity ($\rho$) is...
Ans: Ohm-meter ($\Omega\cdot\text{m}$).
9.Vector form of Ohm's Law is:
Ans: (B) $\vec{J} = \sigma\vec{E}$.
10.Find R for $V = 12\text{ V}$ and $I = 3\text{ A}$.
Ans: $R = \frac{V}{I} = \frac{12}{3} = 4\text{ }\Omega$.
11.Formula relating $R$, $\rho$, $l$, and $A$.
Ans: $R = \rho \frac{l}{A}$.
12.If length and area are both doubled, what happens to R?
Ans: $R_{new} = \rho \frac{2l}{2A} = \rho \frac{l}{A} = R$. The resistance remains unchanged.
13.The average velocity... is called...
Ans: Drift velocity ($v_d$).
14.SI unit of electron mobility ($\mu$) is...
Ans: $\text{m}^2\text{V}^{-1}\text{s}^{-1}$.
15.Relationship between $I$ and $v_d$ is:
Ans: (A) $I = neAv_d$.
16.Define relaxation time ($\tau$).
Ans: It is the average time interval between two successive collisions of free electrons in a conductor.
17.Formula for drift velocity ($v_d$).
Ans: $v_d = \frac{eE\tau}{m}$.
18.Calculate $v_d$ given $\mu = 2.0 \times 10^{-4}$ and $E = 5 \times 10^3$.
Ans: $v_d = \mu E = (2.0 \times 10^{-4}) \times (5 \times 10^3) = 10 \times 10^{-1} = 1\text{ m/s}$.
19.For metals, $\alpha$ is generally...
Ans: Positive (resistivity increases with temperature).
20.For semiconductors, as temp increases, resistivity...
Ans: decreases (negative $\alpha$).
21.Mathematical formula for variation of resistance.
Ans: $R_T = R_0[1 + \alpha(T - T_0)]$ or $R_T = R_0(1 + \alpha\Delta T)$.
22.Match material to resistivity-temperature behavior:
Ans: (a) $\rightarrow$ (ii) increases linearly.
(b) $\rightarrow$ (iii) high and nearly independent.
(c) $\rightarrow$ (i) decreases exponentially.
23.Find $\alpha$ if $R_{27.5} = 2.1$ and $R_{100} = 2.7$.
Ans: $\alpha = \frac{R_2 - R_1}{R_1(T_2 - T_1)} = \frac{2.7 - 2.1}{2.1(100 - 27.5)} = \frac{0.6}{2.1 \times 72.5} \approx 0.0039\text{ }^\circ\text{C}^{-1}$.
24.The commercial unit of electrical energy is...
Ans: Kilowatt-hour ($\text{kWh}$ or "unit").
25.One kilowatt-hour ($1\text{ kWh}$) is exactly equal to...
Ans: $3.6 \times 10^6\text{ Joules}$.
26.Which does NOT represent electrical power ($P$)?
Ans: (B) $IR^2$. (Correct formulas are $VI$, $I^2R$, $V^2/R$).
27.Find Power if $V = 220\text{ V}$ and $I = 0.5\text{ A}$.
Ans: $P = VI = 220 \times 0.5 = 110\text{ Watts}$.
28.What is the resistance of the bulb in Q27?
Ans: $R = \frac{V}{I} = \frac{220}{0.5} = 440\text{ }\Omega$ (or $R = V^2/P = 220^2/110 = 440\text{ }\Omega$).
29.Write Joule's law of heating formula.
Ans: $H = I^2 R t$ (Heat = Current squared $\times$ Resistance $\times$ time).
30.Potential diff when no current is drawn is...
Ans: Electromotive Force (EMF).
31.When discharging, Terminal PD is... EMF.
Ans: less (since $V = E - Ir$).
32.Formula relating $E, V, r, I$ for discharging cell.
Ans: $V = E - Ir$.
33.Identify $E$, $r$, and $R$ in the basic cell circuit.
Ans: $E$ = Electromotive Force of cell, $r$ = Internal resistance of cell, $R$ = External load resistance.
34.Find terminal voltage for $E = 1.5\text{ V}, r = 0.5\text{ }\Omega, I = 1\text{ A}$.
Ans: $V = E - Ir = 1.5 - (1 \times 0.5) = 1.0\text{ V}$.
35.Two identical cells ($E, r$) in series. Equivalent EMF and r?
Ans: Equivalent EMF $E_{eq} = E + E = 2E$. Equivalent internal resistance $r_{eq} = r + r = 2r$.
41.Balanced condition formula for Wheatstone bridge.
Ans: $\frac{P}{Q} = \frac{R}{S}$.
42.A Metre Bridge works on the principle of...
Ans: Balanced Wheatstone Bridge.
43.Current through Galvanometer branch in balanced bridge.
Ans: Exactly zero ($0\text{ A}$).
44.Find $S$ if $P = 10$, $Q = 20$, $R = 30$.
Ans: $\frac{P}{Q} = \frac{R}{S} \implies \frac{10}{20} = \frac{30}{S} \implies S = 30 \times \frac{20}{10} = 60\text{ }\Omega$.
45.Why is Wheatstone unsuitable for very low resistances?
Ans: Because the contact resistances and the resistance of the connecting copper wires become comparable to the low resistance being measured, introducing significant errors.
46.Total length of wire in Metre Bridge.
Ans: $1\text{ meter}$ ($100\text{ cm}$).
47.Potential drop is proportional to...
Ans: Length ($l$) of the wire. ($V \propto l$).
48.Fall of potential per unit length is called...
Ans: Potential Gradient ($k$).
49.Preferred instrument for measuring exact EMF.
Ans: (C) Potentiometer (because it draws no current at null point).
50.Formula to compare EMFs using potentiometer.
Ans: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.
51.Formula for internal resistance using potentiometer.
Ans: $r = R \left( \frac{l_1 - l_2}{l_2} \right)$ or $r = R \left( \frac{l_1}{l_2} - 1 \right)$.
52.Find gradient if EMF $1.5\text{ V}$ balances at $300\text{ cm}$.
Ans: $k = \frac{E}{l} = \frac{1.5\text{ V}}{300\text{ cm}} = 0.005\text{ V/cm}$ (or $0.5\text{ V/m}$).
53.To convert to Ammeter, a very low resistance called a... is connected in...
Ans: Shunt; Parallel.
54.To convert to Voltmeter, a very high resistance is connected in...
Ans: Series.
55.Ideal resistance of perfect Ammeter.
Ans: (B) Zero (so it doesn't affect the circuit current).
56.Ideal resistance of perfect Voltmeter.
Ans: (A) Infinity (so it draws no current from the circuit).
57.Formula for shunt resistance ($S$).
Ans: $S = \frac{I_g \times G}{I - I_g}$.
58.Formula for series resistance ($R$) for voltmeter.
Ans: $R = \frac{V}{I_g} - G$.
59.Calculate shunt for $G=50$, $I_g=2\text{ mA}$, $I=2\text{ A}$.
Ans: $S = \frac{(2 \times 10^{-3}) \times 50}{2 - 0.002} = \frac{0.1}{1.998} \approx 0.05\text{ }\Omega$.
60.Calculate series R for $V=100\text{ V}$ using same G.
Ans: $R = \frac{V}{I_g} - G = \frac{100}{2 \times 10^{-3}} - 50 = 50000 - 50 = 49950\text{ }\Omega$.
